Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 35}{x + 4} = \dfrac{-13x - 71}{x + 4}$
Solution: Multiply both sides by $x + 4$ $ \dfrac{x^2 - 35}{x + 4} (x + 4) = \dfrac{-13x - 71}{x + 4} (x + 4)$ $ x^2 - 35 = -13x - 71$ Subtract $-13x - 71$ from both sides: $ x^2 - 35 - (-13x - 71) = -13x - 71 - (-13x - 71)$ $ x^2 - 35 + 13x + 71 = 0$ $ x^2 + 36 + 13x = 0$ Factor the expression: $ (x + 9)(x + 4) = 0$ Therefore $x = -9$ or $x = -4$ At $x = -4$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -4$, it is an extraneous solution.